Determination of percentage of an unknown salt by titration A volume of 12.80 mL

Question

Determination of percentage of an unknown salt by titration A volume of 12.80 mL of 0.1080 M NaOH solution was used to titrate a 0.587 g sample of unknown containing KHSO.. what is the molecular mass of KHSO? (report answers to 4 or 5 significant figures Submit Answer Tries 0/99 What is the percent by mass of KHSO4 in the unknown? Submit Answer Tries 0/99 In this problem what mass of sample in grams would be needed to deliver about 22.10 mL in the next trial? Submit Answer Tries 0/99 In the second trial above, exactly 1.008 g was transferred into a flask to be titrated. If the initial buret reading is 0.10 mL, predict what the final buret reading be. Submit Answer Tries 0/99

Explanation / Answer

Q1.

MW of KHSO4 = 136.169 g/mol

Q2

mol of base =MV = 0.1080*12.8*10^-3 = 0.0013824

ratio is 1:1

NaOH + KHSO3 = KNaSO3 + H2O

mol of acid, KHSO4 = 0.0013824

mass of KHSO4= mol*MW = 0.0013824*136.169 = 0.1882 g of KHSO4

% mass = 0.1882/0.587*100 = 32.06 %

Q3

mass of sample in grams fo V = 22.10 mL

mass of sample --> 22.1/12.8*0.587 = 1.01349 g of sample

Q4

m = 1.008 g transferred

final V --> (Vf-Vi) = dV

dV = 0.1+22.1 = 22.20 mL

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